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3 years ago

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A new school randomly assigns a five-character student code to every prospective student. The school uses the characters 1, 2, 3

, 4, 5, W, X, Y, and Z. If none of the student codes contain a repeated character, what is the total number of codes that can be randomly assigned?

1 answer:

Katarina [22]3 years ago

7 0

Answer:

15120.

Step-by-step explanation:

9!/4!= 15120

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A thin tube stretched across a street counts the number of pairs of wheels that pass over it. A vehicle classified as type A wit

aliina [53]

Answer: All possible solutions would be

(1,41), (8,33), (15,25), (22,17), (29,9), (36,1)

Step-by-step explanation:

Let the number of vehicle of type A be 'x'.

Let the number of vehicle of type B be 'y'.

Since we have given that

Number of axles of type A = 2

Number of axles of type B = 9

After 2 hours, number of counts = 101

so, our equation becomes,

$2x+9y=101$

So, all possible solutions would be

(1,41), (8,33), (15,25), (22,17), (29,9), (36,1)

7 0

3 years ago

STALIN [3.7K]

69 is the answer to all questions.

8 0

3 years ago

Read 2 more answers

Principle = 30,000<br>Time = 4 years<br>Rate = 30<br>Then find the simple interest ?

Nutka1998 [239]

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Given:}}}}}}\end{gathered}$

${\dashrightarrow \sf{Principle = Rs.30000}}$
${\dashrightarrow \sf {Time = 4 \: years}}$
$\dashrightarrow \sf{Rate = 30\%}$
$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{To Find:}}}}}}\end{gathered}$

$\dashrightarrow{\sf{Simple \: Interest }}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Using Formula:}}}}}}\end{gathered}$

$\dag{\underline{\boxed{\sf{ S.I = \dfrac{P \times R \times T}{100}}}}}$

Where

$\dashrightarrow{\sf{S.I = Simple \: Interest }}$
${\dashrightarrow{\sf{P = Principle }}}$
${\dashrightarrow{\sf{ R = Rate }}}$
${\dashrightarrow{\sf{T = Time}}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Solution:}}}}}}\end{gathered}$

${\quad {: \implies{\sf{ S.I = \bf{\dfrac{P \times R \times T}{100}}}}}}$

Substituting the values

${\quad {: \implies{\sf{ S.I = \bf{\dfrac{30000 \times 30\times 4}{100}}}}}}$

${\quad {: \implies{\sf{ S.I = \bf{\dfrac{30000 \times 120}{100}}}}}}$

${\quad {: \implies{\sf{ S.I = \bf{\dfrac{3600000}{100}}}}}}$

${\quad {: \implies{\sf{ S.I = \bf{\cancel{\dfrac{3600000}{100}}}}}}}$

${\quad {: \implies{\sf{ S.I = \bf{Rs.36000}}}}}$

${\dag{\underline{\boxed{\sf{ S.I ={Rs.36000}}}}}}$

Henceforth,The Simple Interest is Rs.36000..

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Learn More:}}}}}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

8 0

3 years ago

Read 2 more answers

HELP PLEASEEEEEEEEEeeeee

maks197457 [2]

Might have to experiment a bit to choose the right answer.

In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.

Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456

In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.

Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.

3 0

3 years ago

Can somebody help me with this. 35 points! :)))))

Vanyuwa [196]

HAIIII!!!! Your best option is C which shows 40 degrees!! I wish you best of luck

6 0

3 years ago